Question

The oxygen molecule, O2, has a total mass of 5.30×10-26 kg and a rotational inertia of 1.94×10-46 kg-m2 about an axis perpendicular to the center of the line joining the atoms. Suppose that such a molecule in a gas has a speed of 5.96×102 m/s and that its rotational kinetic energy is two-thirds (2/3) of its translational kinetic energy. What then is the molecule's angular speed aboutthe center of mass?

Answer 1

Answer:

$w= 8.0433*10^{12}rad/s$

Explanation:

We need to define the variables.

$m= 5.30*10^{-26}Kg$

$I= 1.94*10^{-46}Kgm^2$

$v= 596m/s$

To use the energy conservative equation we need define $K_r$, that is,

$k_r=\frac{2}{3} k_r$

So,

$\frac{1}{2}Iw^2 = \frac{2}{3} \frac{1}{2} mv^2$

$w^2 = \frac{2mv^2}{3*I}$

$w= \sqrt{\frac{2(5.30*10^{-26})(596)^2}{(3(1.94*10^{-46}))}}$

$w= 8.0433*10^{12}rad/s$